Yeah, that's what they meant.. if you can draw this.. question becomes easy..Btw, the angle is measured from left to the deflected.

Step 2: Draw another diagram

In here, we just shift the deflected vector over..so we can do vector subtraction

Well, impulse = change in momentum. Momentum is a vector, so impulse is also a vector. Therefore, we have to do vector subtraction here.

So, we line them up...impulse = momentum 2 - momentum 1 where the subtraction is vector..But if you noticed.. the mass stays the same, only velocity changes. Since momentum = mass x velocity, it is sufficient to do velocity vector subtraction here. So, in this diagram, we will show only velocity subtraction to simply things. You can do momentum vector subtraction but you have to take the mass along, which is a hassle.

we take the square root of that.. and get 9.793154329 m/s as the answer

but that's change in velocity, not in momentum. change in momentum = change in velocity * mass

which is.. 9.793154329 m/s * 0.401 kg = 3.927054886 kg * m/s

There you have it. change in momentum, which is same as impulse. Enter into CAPA and you should get green light.

Question 7b from CAPA 5

Well, you now know the impulse. Impulse is defined as change in momentum, which is F *D t .

Here is the question:

If the player's head has a mass of 5.30 kg, what is the magnitude of the average acceleration of the player's head during the impact? Assume that over the brief time of the impact, 28.80 ms, the player's head can be treated separately from the player's body.

What is the speed of the car after all the cannon balls have come to rest on the right side?

This is a rather conceptual question: Here is what happens inside the railroad car.

The cannon shoots cannon balls. It exherts a force on the cannon ball. According to Newton's third law, the cannon ball will exhert a equal in magnitude but opposite in direction force back. This reaction force will push on the cannon, which is fixed in the railroad cart. This causes the entire cart to move. When the ball hits on the other side of the cart, it exherts a force on the car that is in the opposite direction to the car's movement. This effectively halt's cart's movement. (Since the cart's movement is caused by the same force that propels the cannon ball forward, these two are equal and therefore cancel each other out)

So, when all the cannonballs are shoot and hit the right side of the cart, the cart is stationary and have a velocity of 0 m/s.

Here are two pictures demonstrating the concepts.

Question 9 from CAPA 5

During a violent thunderstorm, hail the size of marbles (diameter = 1.50 cm) falls at a speed of 15.0 m/s. There are estimated to be 205.0 hailstones per cubic meter of air. Assume that, as for ice, 1.00 cm3 of hail has a mass of 0.920 g. What is the mass of each hailstone?

Ok, for this question, forget about the 205 hailstones /air!! It will mislead you. You have to caculate the # yourself. (The reason is that they tell you 0.920 g per cm^3 of ice, which is not air!)

First, find the volume of the hail. The volume is 4/3pr³, which in my case.. it is

4/3 * pi * (1.50/2 /100)³, the reason for the /100 is to convert to meters.

The answer is... 0.000001767 m³.

Use 1 m³ divide by 0.000001767 m³ to find out how many hails per m³. Answer for me is 565930.9564 hail stones.

Then they tell you that there are 0.920 g per cm³, which means that 0.920 x 1. * 10^6 will give you the mass in m³, which is 920,000 g. Use 920,000 g to divide by 565930.9564 hailstones to find how mass per hail stone.. 1.62564 g is my answer.

Question 9b from CAPA 5

Yes! Last question!

Ok, here is the question:

What is the magnitude of the force exerted by the hail on a 8.0 m x 12.0 m flat roof during the storm ? Ignore the bounce of the hail on impact.

Step 1: Volume

Lets look at what happens every second. The hailstones falls at 15 m/s. This means that every second, there would be 15 m of hailstones stacked vertically. This gave us the height.

We are also given the area of the roof, which is 8.0 m x 12.0 m. Therefore, we can use 15 m * 8.0 m * 12.0 m to get the volume of hailstone that hits the roof every second.

This give us 1440 m³.

Step 2: Find the number of hailstones in that volume

Now, we know there are 205 hailstones per volume of air. Thus, to find the total mass of the hailstones failing on the roof every second, we have to find the # of hailstones in that volume first.

Since that volume is volume of air, we can safely use the 205 halstones/volume of air they tell us. This results in 205 / m³* 1440 m³ = 295,200 hailstones that hit the roof every second.

Step 3: Find the mass of hailstones that hit roof /second

We know there are 295,200 hailstones that hit the roof per second and we know that each stone weighs 1.62564 g. (YOU MUST NEVER USE THE ANSWER FROM 9a THAT THEY GIVE YOU, IT IS ROUNDED AND THUS WILL SCREW YOU. USE THE ORIGINAL UNROUNDED VALUE)

Thus, 295,200 * 1.62564 g = 479888.928 g, which is 479.888928 kg.

Step 4: Find the momentum of the hailstones that hit the roof every second

Momentum = mass * velocity, you know both. (Momentum of the all the hailstones that hit the roof every second)

p = 479.888928 kg * 15 m/s = 7198.33392 kg *m/s

Step 5: Find the force

Force = momentum / change in time. Change in time is 1s. so you have 7198.33392 kg *m/s divide by 1s which will give you 7198.33392 N. Enter that and you are done!